# Nonlocal Conditions for Lower Semicontinuous Parabolic Inclusions

- Abdelkader Boucherif
^{1}Email author

**2011**:109570

https://doi.org/10.1155/2011/109570

© Abdelkader Boucherif. 2011

**Received: **4 December 2010

**Accepted: **11 February 2011

**Published: **9 March 2011

## Abstract

We discuss conditions for the existence of at least one solution of a discontinuous parabolic equation with lower semicontinuous right hand side and a nonlocal initial condition of integral type. Our technique is based on fixed point theorems for multivalued maps.

## 1. Introduction

Let be an open bounded domain in , , with a smooth boundary . We denote the norm (usually the Euclidean norm) of by . Let be a positive real number. Set and . For we denote its partial derivatives (when they exist) by .

we have the following classical result.

Assume that the function is Hölder continuous on and is continuous on . Then problem (1.3), (1.4), (1.8) has a unique solution , which for each , is given by

- (i)
- (ii)
- (iii)
- (iv)
- (v)
- (vi)
, for some positive constants (see [2]).

Since , it is clear that the functions and are continuous. Let and let Also, property (vi) above shows that .

Parabolic problems with discontinuous nonlinearities arise as simplified models in the description of porous medium combustion [5], chemical reactor theory [6]. Also, best response dynamics arising in game theory can be modeled by a parabolic equation with a discontinuous right hand side [7, 8]. Parabolic problems with discontinuous nonlinearities have been also investigated in the papers [9–13]. On the other hand, parabolic problems with integral boundary conditions appear in the modeling of concrete problems, such as heat conduction [14, 15] and thermoelasticity [16]. Also, the importance of nonlocal conditions and their applications in different field has been discussed in [17, 18]. Several papers have been devoted to the study of parabolic problems with integral conditions [19, 20]. Next, we state some important facts about multivalued functions and results that will be used in the remainder of the paper.

A subset is measurable if belongs to the -algebra generated by all sets of the form where is Lebesgue measurable in and is Borel measurable in . Let and be Banach spaces. denotes the set of all nonempty subsets of . The domain of a multivalued map is the set Dom( has closed values if is a closed subset of for each and we write . Also, denotes the set of all nonempty closed and convex subsets of . is bounded if is called lower semicontinuous (lsc) on if is open in whenever is open in , or the set is closed in whenever is closed in . For more details on multivalued maps, we refer the interested reader to the books [21–24].

Let denote the Kuratowski measure of noncompactness. See [25] for definitions and details.

Theorem 1.2 (see [26, Theorem 3.1]).

Let be a separable Banach space. Assume the following conditions hold. There exists , independent of , with for any solution to a.e. on for each is a closed map, is a bounded subset of , and for all with strict inequality if . Then the inclusion has a solution .

## 2. Main Result

By a solution of problem (1.10), (7), (8) we mean a function such that there exists a function with for each and (1.3), (1.4), (1.5) hold.

Theorem 2.1.

Assume that the following conditions are satisfied.

(HF) is measurable, is lsc for a.e. , there exist such that with 2Vol and there exists such that for any bounded set ,

(Hk) is continuous, bounded and there exists such that .

Then problem (1.10), (7), (8) has a solution provided that .

Proof.

Let . Then is nonempty, closed, and bounded subset of .

Since the multifunction has nonempty, closed and convex values, it follows that has nonempty, closed, and convex values. Since is a continuous single valued operator, it is clear that has nonempty, closed, and convex values. Next, we can easily show that is a closed map (i.e., has a closed graph) and is a bounded subset of .

This shows that is a condensing multivalued map.

By Theorem 3.1 in [26], has a fixed point in , which is a solution of problem (1.10), (7), (8). This completes the proof of the main result.

## Declarations

### Acknowledgments

This work is part of an ongoing research project FT090001. The author is grateful to KFUPM for its constant support. The author would like to thank an anonymous referee for his/her comments.

## Authors’ Affiliations

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